Daily Practice Questions Questions for CAT with Answers PDF

The last digit in the product can be one of the required digits, if and only if, the last digit of each of the numbers is one of the numbers 1, 3, 7 or 9.
Probability of choosing each of the four numbers is 4/10.
As the last digit of required numbers can be chosen in four different ways and total number of ways of choosing a digit at units place is 10,

Let 'x' be the probability, that card labelled '1' is selected.
Probability that card labelled '2' is selected = 2x
Probability that card labelled '3' is selected = 3x and so on
Now, x + 2x + 3x + … + 100x = 1, which means x = 1/5050

Required probability = P (drawing a non-ace each of the first four times and drawing an ace the fifth time) =

A leap year has 366 days, or 52 weeks + 2 days. This means that 52 Sundays will be there, but for the 53rd, we have to evaluate the possibility of the two extra or odd days to fall on Sunday or Monday.
The possible outcomes for two odd days can be (Mon, Tue); (Tue, Wed); (Wed, Thu); (Thu, Fri); (Fri, Sat); (Sat, Sun) and (Sun, Mon).
Number of favourable cases = 3
Required probability = 3/7

Probability that the candidate does not get an offer from the first interview is 4/5.
Probability that the candidate does not get an offer from the second interview is 5/6.
Total probability of not getting an offer 

Hence, probability of getting at least one job 

Probability of drawing a Queen = 4/52 = 1/13
Since the first card is replaced, the pack will again have 52 cards. So, the probability of drawing a King = 4/52 = 1/13
Both the events are independent, hence the probability of drawing both cards in succession 

Probability that one head occurs

Probability that all tails occur 

Hence, probability that at least two heads occur 

The total number of ways of choosing two numbers out of 1, 2, 3 ........, 30 is
³⁰C₂ = 435
⇒ Exhaustive number of cases = 435
Since a² - b² is divisible by 3 if either a and b both are divisible by 3 or none of a and b is divisible by 3, so the favourable number of cases = ¹⁰C₂ + ²⁰C₂ = 235
Hence, the required probability = 235/435 = 47/87

S = sample space. n(s) = No. of ways of drawing two boxes out of (5 + 7) boxes = ¹²C₂ = (12 × 11)/(2 × 1) = 66.
E = Event of getting both boxes of the same shape.
n(E) = No. of ways of drawing (2 boxes out of 5 or 2 boxes out of 7) = ⁵C₂ + ⁷C₂ = (5 × 4)/(2 × 1) + (7 × 6)/(2 × 1) = 31.
P(E) = n(E)/n(S) = 31/66.

S = Sample space = 52.
There are 13 hearts including 1 seven.
There are 3 more sevens in spade, club and diamond.
E = event of getting a seven or a heart.
n(E) = 13 + 3 = 16.
P(E) = n(E)/n(S) = 16/52 = 4/13

Here, the sample space is 52.
Probability that the first card is a queen = 4/52 = 1/13
Probability that the second card is a queen = 3/51
Hence, probability that the cards drawn are queens 

A: The first dice shows 4, 5 or 6 dots.
B: The seconds dice shows 5 or 6 dots.
P(A) = 3/6 = 1/2
P(B) = 2/6 = 1/3
So, the required probability is
 

S = Sample space and E = event of choosing 1 woman and 2 men.
n(S) = Number of ways of selecting 3 people out of 25

n(E) = (¹⁰C₁ × ¹⁵C₂) = 1050.
Therefore, P(E) = n(E)/n(S) = 1050/2300 = 21/46

Total number of balls = 5 + 9 + 6 = 20.
E = event that the ball drawn is neither red nor green = event that the ball drawn is black = 6.
Therefore, P(E) = 6/20.

Elementary events associated to the random experiment of throwing two dice are: 
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4),(4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total no. of events = 36
Let A be the event of getting an even number on one and a multiple of 3 on the other.
A : (2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)
So, favourable no. of events = 11
Hence, probability = (No. of favourable events)/(Total no. of events)
Probability = 11/36

Total number of balls = 4 + 3 + 5 = 12
Number of balls which are either red or yellow = 3 + 4 = 7
Here, the total sample space is 12 and the event value is 7.
Hence, probability of getting a red or a yellow ball = 7/12

There are 3 cases possible as follows:
Case 1:
The red die throws up a 2 and the green die throws up a 2 or a 1, the probability of which is (1/6)(2/6) = 2/36 = 1/18

Case 2:
The red die throws up a 3 and the green die throws up a 3 or a 1, the probability of which is (1/6)(2/6) = 2/36 = 1/18

Case 3:
The red die throws up a 5 and the green die throws up a 5 or a 1, the probability of which is (1/6)(2/6) = 2/36 = 1/18
Thus, the required probability = (1/18) + (1/18) + (1/18) = 3 × (1/18) = 1/6

The possible outcomes are:
(1, 1); (1, 2); (2, 1), (2, 2); (3, 1); (1, 3).
Out of six cases, in two cases there is exactly one ‘2’
Thus, the correct answer is 2/6 = 1/3.

The event can be defined as:
First bag is selected and red ball is drawn.
or second bag is selected and red ball is drawn.
1/2 X 5/12 + / X 3/15 = (5/24) + (3/30) = 37/120

One head and seven tails would have eight positions where the head can come.
Thus, 8 x (1/2)⁸ = (1/32)